How can I reorder a list? [closed]

Given an arbitrary array of size n , I'd like to reorganize the elements of the array based on the array's discrete indices. Python example:

# Unique array of size n [ "a", "b", "c", "d", "e", . ] # Indices of array [ 0, 1, 2, 3, 4, . ] # Desired re-organization function 'indexMove' indexMove( [ "a", "b", "c", "d", "e", . ], [ , 4, 0, 2, 3, . 1 ] ) # Desired output from indexMove operation [ , "e", "a", "c", "d", . "b" ] 

Without specifying how you want to order the items its hard to answer. Do you want to sort them? Shuffle them? Remove some of them?

@tvanfosson: In this case, arbitrary could also mean: take an arbitrary (but well defined) sort function.

@SilentGhost It will have a new index. May be 4. The point is that I know the new order of the items.

12 Answers 12

You can do it like this

mylist = ['a', 'b', 'c', 'd', 'e'] myorder = [3, 2, 0, 1, 4] mylist = [mylist[i] for i in myorder] print(mylist) # prints: ['d', 'c', 'a', 'b', 'e'] 

@Adam Thanks! What's the time and space complexity of this operation? Does it iterate over mylist when assigning entries from [mylist[i] for i in myorder] to mylist[:] ?

Hey, I am new to Python and I am curious how does this line of code "[mylist[i] for i in myorder]" work? I have only learnt the basic for loop.

Wow, oldie but a goodie. Thanks for taking me back @ZionAdams. Check out List Comprehensions: realpython.com/list-comprehension-python/…

>>> a = [1, 2, 3] >>> a[0], a[2] = a[2], a[0] >>> a [3, 2, 1] 

>>> import random >>> x = [1,2,3,4,5] >>> random.shuffle(x) >>> x [5, 2, 4, 3, 1] 

no error, I just type: 1) import random; x = [1, 2, 3]; random.shuffle(x); #it returns None; and 2) I tried np.random.shuffle. the results are the same.

Oh, sorry! I did not realize that I directly change the value of x. It did return None. And it works. Thanks.

Is the final order defined by a list of indices ?

>>> items = [1, None, "chicken", int] >>> order = [3, 0, 1, 2] >>> ordered_list = [items[i] for i in order] >>> ordered_list [, 1, None, 'chicken'] 

>>> a=["a","b","c","d","e"] >>> a[0],a[3] = a[3],a[0] >>> a ['d', 'b', 'c', 'a', 'e'] 

You can provide your own sort function to list.sort() :

• cmp specifies a custom comparison function of two arguments (list items) which should return a negative, zero or positive number depending on whether the first argument is considered smaller than, equal to, or larger than the second argument: cmp=lambda x,y: cmp(x.lower(), y.lower()) . The default value is None .
• key specifies a function of one argument that is used to extract a comparison key from each list element: key=str.lower . The default value is None .
• reverse is a boolean value. If set to True, then the list elements are sorted as if each comparison were reversed.

@SilentGhost: This is meant as a general answer. In the OPs case your answer is more appropriate. Nevertheless I think it is important to know that a generic solution exists.

If you use numpy there's a neat way to do it:

items = np.array(["a","b","c","d"]) indices = np.arange(items.shape[0]) np.random.shuffle(indices) print(indices) print(items[indices]) 

This code returns:

[1 3 2 0] ['b' 'd' 'c' 'a'] 

If you do not care so much about efficiency, you could rely on numpy's array indexing to make it elegant:

a = ['123', 'abc', 456] order = [2, 0, 1] a2 = list( np.array(a, dtype=object)[order] ) 

One more thing which can be considered is the other interpretation as pointed out by darkless

Code in Python 2.7

1. Reorder by value - Already solved by AJ above
2. Reorder by index

mylist = ['a', 'b', 'c', 'd', 'e'] myorder = [3, 2, 0, 1, 4] mylist = sorted(zip(mylist, myorder), key=lambda x: x[1]) print [item[0] for item in mylist] 

This will print ['c', 'd', 'b', 'a', 'e']

From what I understand of your question, it appears that you want to apply a permutation that you specify on a list . This is done by specifying another list (lets call it p ) that holds the indices of the elements of the original list that should appear in the permuted list . You then use p to make a new list by simply substituting the element at each position by that whose index is in that position in p .

def apply_permutation(lst, p): return [lst[x] for x in p] arr=list("abcde") new_order=[3,2,0,1,4] print apply_permutation(arr,new_order) 

This prints ['d', 'c', 'a', 'b', 'e'] .

This actually creates a new list , but it can be trivially modified to permute the original "in place".